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 long long N = 2e5 + 100, MOD = 1e9 + 7, INF = 1e15, SQ = 350, LOG = 20;
int n, m, x1, x2, z, t, y2, maxi, maxi2, mini, ptr, a[N], b[N], p, q, k, sum1,
    sum2, x, y, dp[N], pre, ans;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  dp[1] = -1;
  for (int i = 2; i <= n; i++) {
    if (a[i - 1] != a[i]) {
      dp[i] = i - 1;
    } else {
      dp[i] = dp[i - 1];
    }
  }
  while (m--) {
    ans = -1;
    scanf("%d %d %d", &x, &y, &z);
    if (a[y] == z) {
      if (a[x] != z) {
        ans = x;
      } else {
        if (dp[y] == dp[x]) {
          ans = -1;
        } else {
          ans = dp[y];
        }
      }
    } else {
      ans = y;
    }
    printf("%d\n", ans);
  }
}