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 N = 2000050;
int n, m, A[N], l, r, x, dp[N];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < (int)(n); i++) {
    scanf("%d", &A[i]);
  }
  for (int i = (n)-1; i >= 0; i--) {
    if (i + 1 == n) {
      dp[i] = -1;
    } else {
      if (A[i] != A[i + 1]) {
        dp[i] = i + 1;
      } else {
        dp[i] = dp[i + 1];
      }
    }
  }
  while (m--) {
    scanf("%d %d %d", &l, &r, &x);
    --l;
    --r;
    if (A[l] != x) {
      printf("%d\n", l + 1);
    } else {
      int v = dp[l];
      if (v > r || v == -1) {
        printf("-1\n");
      } else {
        printf("%d\n", v + 1);
      }
    }
  }
  return 0;
}