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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int i, j, n, m, p, v, a[100005];
  long long int ans = 0;
  set<int> npres[17];
  for (i = 0; i < 17; i++) npres[i].insert(0);
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    for (j = 0; j < 17; j++)
      if (a[i] & 1 << j)
        ans += (i - (*(npres[j].rbegin()))) * (1ll << j);
      else
        npres[j].insert(i);
  }
  for (i = 0; i < 17; i++) npres[i].insert(n + 1);
  for (i = 0; i < m; i++) {
    cin >> p >> v;
    for (j = 0; j < 17; j++)
      if (a[p] & 1 << j) {
        auto itr = npres[j].lower_bound(p);
        ans -= (1ll * (*itr - p)) * (p - *(--itr)) * (1ll << j);
      } else
        npres[j].erase(p);
    a[p] = v;
    for (j = 0; j < 17; j++)
      if (a[p] & 1 << j) {
        auto itr = npres[j].lower_bound(p);
        ans += (1ll * (*itr - p)) * (p - *(--itr)) * (1ll << j);
      } else
        npres[j].insert(p);
    cout << ans << "\n";
  }
}