Problem: You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi.

For the i-th query find any position pi (li ≤ pi ≤ ri) so that api ≠ xi.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the number of elements in a and the number of queries.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Each of the next m lines contains three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106) — the parameters of the i-th query.

Output

Print m lines. On the i-th line print integer pi — the position of any number not equal to xi in segment [li, ri] or the value  - 1 if there is no such number.

Examples

Input

6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2


Output

2
6
-1
4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
struct segtree {
  int node[maxn << 2];
  void pushup(int rt) {
    node[rt] = node[rt << 1] == node[rt << 1 | 1] ? node[rt << 1] : -1;
  }
  void build(int rt, int l, int r) {
    if (l == r) {
      scanf("%d", node + rt);
      return;
    }
    int m = (l + r) >> 1;
    build(rt << 1, l, m);
    build(rt << 1 | 1, m + 1, r);
    pushup(rt);
  }
  int query(int rt, int l, int r, int x, int y, int c) {
    if (~node[rt] && node[rt] == c) return -1;
    if (l == r) return l;
    int m = (l + r) >> 1, ret = -1;
    if (x <= m) ret = query(rt << 1, l, m, x, y, c);
    if (m < y && ret == -1) ret = query(rt << 1 | 1, m + 1, r, x, y, c);
    return ret;
  }
} tree;
int main() {
  int n, q;
  while (~scanf("%d%d", &n, &q)) {
    tree.build(1, 1, n);
    while (q--) {
      int l, r, c;
      scanf("%d%d%d", &l, &r, &c);
      printf("%d\n", tree.query(1, 1, n, l, r, c));
    }
  }
  return 0;
}