Problem: Inna is fed up with jokes about female logic. So she started using binary logic instead.

Inna has an array of n elements a1[1], a1[2], ..., a1[n]. Girl likes to train in her binary logic, so she does an exercise consisting of n stages: on the first stage Inna writes out all numbers from array a1, on the i-th (i ≥ 2) stage girl writes all elements of array ai, which consists of n - i + 1 integers; the k-th integer of array ai is defined as follows: ai[k] = ai - 1[k] AND ai - 1[k + 1]. Here AND is bit-wise binary logical operation.

Dima decided to check Inna's skill. He asks Inna to change array, perform the exercise and say the sum of all <image> elements she wrote out during the current exercise.

Help Inna to answer the questions!

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — size of array a1 and number of Dima's questions. Next line contains n integers a1[1], a1[2], ..., a1[n] (0 ≤ ai ≤ 105) — initial array elements.

Each of next m lines contains two integers — Dima's question description. Each question consists of two integers pi, vi (1 ≤ pi ≤ n; 0 ≤ vi ≤ 105). For this question Inna should make a1[pi] equals vi, and then perform the exercise. Please, note that changes are saved from question to question.

Output

For each question print Inna's answer on a single line.

Examples

Input

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


Output

6
4
7
12

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m;
long long ans, sum[100009];
bool bit[100009][18];
int main() {
  set<int> z[18];
  cin >> n >> m;
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + i;
  int h[18];
  memset(h, 0, sizeof(h));
  for (int i = 0, k; i < n; i++) {
    cin >> k;
    for (int j = 0; j < 18; j++) {
      if (1 << j & k) {
        bit[i][j] = true;
        h[j]++;
      } else {
        ans += sum[h[j]] * (1 << j);
        h[j] = 0;
        z[j].insert(i);
      }
    }
  }
  for (int i = 0; i < 18; i++) {
    if (h[i] > 0) ans += sum[h[i]] * (1 << i);
  }
  for (int i = 0, p, v; i < m; i++) {
    cin >> p >> v;
    p--;
    for (int j = 0, l, r; j < 18; j++) {
      bool b = ((1 << j & v) > 0);
      if (bit[p][j] == b)
        continue;
      else {
        bit[p][j] = b;
        if (b == true) {
          set<int>::iterator i = z[j].lower_bound(p);
          set<int>::iterator k = i;
          if (i == z[j].begin())
            l = -1;
          else {
            k--;
            l = *k;
          }
          k = i;
          i++;
          if (i == z[j].end())
            r = n;
          else
            r = *i;
          z[j].erase(k);
          long long ad = sum[r - l - 1] - sum[p - l - 1] - sum[r - p - 1];
          ans = ans + ad * (1 << j);
        } else {
          set<int>::iterator i = z[j].lower_bound(p);
          set<int>::iterator k = i;
          if (i == z[j].end())
            r = n;
          else
            r = *i;
          if (i == z[j].begin())
            l = -1;
          else {
            k--;
            l = *k;
          }
          z[j].insert(p);
          long long ad = sum[p - l - 1] + sum[r - p - 1] - sum[r - l - 1];
          ans = ans + ad * (1 << j);
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}