给定一个长度为$n(1\le \mid s\mid \le 10^5)$的只由小写字母构成的字符串$s$，有$m,(1\le m\le 10^6)$个询问：

每次询问给定$l,r$，问$s$是否存在一个字串$t$，满足$\mid t\mid <2(r-l+1)$，且$s[l,r]$在$t$中出现最少两次。

让$s[l,r]$在$t$上匹配，假设匹配的结束位置为$p_1,p_2,\dots$，如果$t$是一个合法串，则一定存在两个不同的数$i,j$，

且满足$p_i\ne p_{j}ANDabs(p_i-p_j)<r-l+1$，从这一点出发，我们考虑后缀自动机。

首先我们构建$SAM$，建出$parent$树，我们不难找到$endpos=r$的节点，考虑从这个点向上跳，

当我们跳到最上面的满足$len\ge r-l+1$的节点时，显然这个点所代表的$endpos$集合中的字串，一定都含有$s[l,r]$这个后缀。

简单的理解一下也就是$s[l,r]$可以在这些节点匹配上，所以我们只要判断这些节点是否存在两个$endpos$满足其差值$\le r-l+1$即可。

这里我用的是线段树合并，对线段树上的节点记录其区间最左边有值的树，区间最有边有值的树，

以及当前区间的答案，然后合并上去即可，我们只要记录下每个$endpos$集合里面的最小的答案即可，整体复杂度$T\times O(n\log n+m)$。

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10;int num, last, endpos[N << 1];int head[N << 1], nex[N << 1], to[N << 1], cnt;int fa[N << 1][21], ans[N << 1], idx[N], n, m;int root[N << 1], ls[N << 6], rs[N << 6], tot;char str[N];struct Res {int l, r, minn;
}a[N << 6];struct statu {int len, link, next[26];
}s[N << 1];void push_up(int rt) {if (ls[rt] && rs[rt]) {a[rt] = {a[ls[rt]].l, a[rs[rt]].r, min({a[ls[rt]].minn, a[rs[rt]].minn, a[rs[rt]].l - a[ls[rt]].r})};}else if (ls[rt]) {a[rt] = a[ls[rt]];}else {a[rt] = a[rs[rt]];}
}void update(int &rt, int l, int r, int x) {if (!rt) {rt = ++tot;}if (l == r) {a[rt] = {x, x, 0x3f3f3f3f};return ;}int mid = l + r >> 1;if (x <= mid) {update(ls[rt], l, mid, x);}else {update(rs[rt], mid + 1, r, x);}push_up(rt);
}int merge(int x, int y, int l, int r) {if (!x || !y) {return x | y;}if (l == r) {return x;}int mid = l + r >> 1;ls[x] = merge(ls[x], ls[y], l, mid);rs[x] = merge(rs[x], rs[y], mid + 1, r);push_up(x);return x;
}void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void Init() {for (int i = 0; i < num; i++) {s[i].len = s[i].link = endpos[i] = head[i] = root[i] = 0;for (int j = 0; j < 26; j++) {s[i].next[j] = 0;}}num = last = 0;for (int i = 1; i <= tot; i++) {ls[i] = rs[i] = 0;}tot = 0;
}void init() {s[0].len = 0, s[0].link = -1;num++, last = 0, cnt = 1;
}void extend(int c, int id) {int cur = num++;s[cur].len = s[last].len + 1, endpos[cur] = id, idx[id] = cur;int p = last;while (p != -1 && !s[p].next[c]) {s[p].next[c] = cur;p = s[p].link;}if (p == -1) {s[cur].link = 0;}else {int q = s[p].next[c];if (s[p].len + 1 == s[q].len) {s[cur].link = q;}else {int clone = num++;s[clone] = s[q];s[clone].len = s[p].len + 1;while (p != -1 && s[p].next[c] == q) {s[p].next[c] = clone;p = s[p].link;}s[q].link = s[cur].link = clone;}}last = cur;
}void dfs(int rt, int f) {fa[rt][0] = f;for (int i = 1; i <= 20; i++) {fa[rt][i] = fa[fa[rt][i - 1]][i - 1];}for (int i = head[rt]; i; i = nex[i]) {if (to[i] == f) {continue;}dfs(to[i], rt);root[rt] = merge(root[rt], root[to[i]], 1, n);}if (endpos[rt]) {update(root[rt], 1, n, endpos[rt]);}ans[rt] = a[root[rt]].minn;
}void build() {for (int i = 1; i < num; i++) {add(s[i].link, i);}dfs(0, -1);
}int get(int rt, int len) {for (int i = 20; i >= 0; i--) {if (fa[rt][i] != 0 && fa[rt][i] != -1 && s[fa[rt][i]].len >= len) {rt = fa[rt][i];}}return rt;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);int T;scanf("%d", &T);while (T--) {scanf("%d %d %s", &n, &m, str + 1);init();for (int i = 1; i <= n; i++) {extend(str[i] - 'a', i);}build();for (int i = 1, l, r; i <= m; i++) {scanf("%d %d", &l, &r);int rt = get(idx[r], r - l + 1);if (ans[rt] < r - l + 1) {puts("Yes");}else {puts("No");}}Init();}return 0;
}