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;
inline void in(int &n) {
  n = 0;
  int ch = getchar();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar();
  }
  n = n * sign;
}
int a[200005];
vector<int> v[1000006];
int main() {
  int n, m;
  in(n);
  in(m);
  for (int i = 1; i <= n; i++) {
    in(a[i]);
    v[a[i]].push_back(i);
  }
  while (m--) {
    int l, r, x;
    in(l);
    in(r);
    in(x);
    if (l == r) {
      if (a[l] == x) {
        printf("-1\n");
      } else {
        printf("%d\n", l);
      }
      continue;
    }
    if (a[l] != x) {
      printf("%d\n", l);
      continue;
    }
    int pos1 = lower_bound(v[x].begin(), v[x].end(), l) - v[x].begin();
    int low = l + 1;
    int up = r;
    while (low < up) {
      int mid = low + ((up - low) >> 1);
      if (a[mid] != x) {
        up = mid;
        continue;
      }
      int pos2 = lower_bound(v[x].begin(), v[x].end(), mid) - v[x].begin();
      if (pos2 - pos1 == mid - l) {
        low = mid + 1;
      } else if (pos2 - pos1 < mid - l) {
        up = mid;
      }
    }
    if (a[low] == x) {
      printf("-1\n");
    } else {
      printf("%d\n", low);
    }
  }
}