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 = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
int a[maxn], pre[maxn];
int find(int x) {
  if (x != pre[x])
    return pre[x] = find(pre[x]);
  else
    return x;
}
void join(int x, int y) {
  int fx = find(x);
  int fy = find(y);
  if (fx != fy) {
    pre[fx] = fy;
  }
  return;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    pre[i] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == a[i - 1]) join(i, i - 1);
  }
  while (m--) {
    int l, r, x;
    scanf("%d %d %d", &l, &r, &x);
    int flag = 0;
    while (r >= l) {
      if (a[r] != x) {
        printf("%d\n", r);
        flag = 1;
        break;
      }
      r = find(r) - 1;
    }
    if (!flag) printf("-1\n");
  }
  return 0;
}