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;
void err(istream_iterator<string> it) {
  cout << '\n';
  ;
}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ", ";
  err(++it, args...);
}
template <typename T>
inline void Int(T &n) {
  n = 0;
  int f = 1;
  register int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
  n = n * f;
}
template <typename T, typename TT>
inline void Int(T &n, TT &m) {
  Int(n);
  Int(m);
}
template <typename T, typename TT, typename TTT>
inline void Int(T &n, TT &m, TTT &l) {
  Int(n, m);
  Int(l);
}
double ind() {
  double x;
  scanf("%lf", &x);
  return x;
}
int set_1(int n, int pos) { return n = (n | (1 << pos)); }
int reset_0(int n, int pos) { return n = n & ~(1 << pos); }
bool check_bit(int n, int pos) { return n = n & (1 << pos); }
int dx[8] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[8] = {0, 1, 0, -1, -1, 1, -1, 1};
const int N = (int)1e6 + 5;
const int M = (int)1e9 + 7;
const double pi = 2 * acos(0.0);
const double eps = 1e-9;
int a[N], last[N];
int main() {
  int t = 1, te = 0;
  while (t--) {
    int n, m;
    Int(n, m);
    for (int i = 1; i <= n; ++i) Int(a[i]);
    last[0] = -1;
    for (int i = 1; i <= n; ++i) {
      if (a[i] != a[i - 1])
        last[i] = i - 1;
      else {
        last[i] = last[i - 1];
      }
    }
    for (int i = 1; i <= m; ++i) {
      int l, r, x;
      Int(l, r, x);
      if (a[r] != x)
        printf("%d\n", r);
      else if (last[r] >= l)
        printf("%d\n", last[r]);
      else {
        printf("-1\n");
      }
    }
  }
  return 0;
}