Problem: Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and number y, GukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to  - 1. GukiZ also prepared a problem for you. This time, you have two types of queries: 

  1. First type has form 1 l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x. 
  2. Second type has form 2 y and asks you to find value of GukiZiana(a, y). 



For each query of type 2, print the answer and make GukiZ happy!

Input

The first line contains two integers n, q (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104), size of array a, and the number of queries. 

The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109), forming an array a. 

Each of next q lines contain either four or two numbers, as described in statement:

If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109), first type query.

If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 109), second type query.

Output

For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.

Examples

Input

4 3
1 2 3 4
1 1 2 1
1 1 1 1
2 3


Output

2


Input

2 3
1 2
1 2 2 1
2 3
2 4


Output

0
-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long n, q, k;
long long A[500010], add[2010];
vector<pair<long long, long long> > V[2010];
inline void update(long long x, long long l, long long r, long long v) {
  if (r - l + 1 == k) {
    add[x] += v;
    return;
  }
  V[x].clear();
  for (long long i = (x - 1) * k + 1; i <= min(n, x * k); i++)
    if (i < l || i > r) {
      A[i] += add[x];
      V[x].push_back(make_pair(A[i], i));
    } else {
      A[i] += add[x] + v;
      V[x].push_back(make_pair(A[i], i));
    }
  add[x] = 0;
  sort(V[x].begin(), V[x].end());
}
int main() {
  cin >> n >> q;
  k = 707;
  for (long long i = 1; i <= n; i++) cin >> A[i];
  for (long long i = 1; i <= n; i += k) {
    for (long long j = i; j <= min(n, i + k - 1); j++)
      V[i / k + 1].push_back(make_pair(A[j], j));
    sort(V[i / k + 1].begin(), V[i / k + 1].end());
  }
  for (long long i = 1; i <= q; i++) {
    long long type, l, r, x, bek;
    cin >> type;
    if (type == 1) {
      cin >> l >> r >> x;
      for (long long j = 1; j <= r; j += k)
        if (j + k - 1 >= l) update(j / k + 1, max(l, j), min(r, j + k - 1), x);
    } else {
      l = r = bek = 0;
      cin >> x;
      for (long long j = 1; j <= n; j += k) {
        long long w = x - add[j / k + 1];
        long long a = lower_bound(V[j / k + 1].begin(), V[j / k + 1].end(),
                                  make_pair(w, 0LL)) -
                      V[j / k + 1].begin();
        if (!bek and a != V[j / k + 1].size() and V[j / k + 1][a].first == w) {
          l = V[j / k + 1][a].second;
          bek = 1;
        }
        w = x - add[j / k + 1];
        long long b = lower_bound(V[j / k + 1].begin(), V[j / k + 1].end(),
                                  make_pair(w + 1, 0LL)) -
                      V[j / k + 1].begin() - 1;
        if (b != -1 and V[j / k + 1][b].first == w) r = V[j / k + 1][b].second;
      }
      if (r)
        cout << r - l << endl;
      else
        cout << -1 << endl;
    }
  }
}