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;
const int N = 1e5 + 10;
int a[N];
set<int> z[20];
long long sum[20];
void one(int p, int v) {
  z[p].erase(v);
  set<int>::iterator low = --z[p].lower_bound(v);
  set<int>::iterator up = z[p].upper_bound(v);
  sum[p] += 1ll * (v - *low) * (*up - v);
}
void zero(int p, int v) {
  z[p].insert(v);
  set<int>::iterator low = --z[p].lower_bound(v);
  set<int>::iterator up = z[p].upper_bound(v);
  sum[p] -= 1ll * (v - *low) * (*up - v);
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < 20; i++)
    for (int j = 0; j <= n + 1; j++) z[i].insert(j);
  for (int i = 0; i < 20; i++)
    for (int j = 0; j <= n + 1; j++)
      if (a[j] & (1 << i)) one(i, j);
  while (m--) {
    int p, v;
    scanf("%d%d", &p, &v);
    long long ans = 0;
    for (int i = 0; i < 20; i++) {
      if (v & (1 << i)) {
        if (!z[i].count(p)) continue;
        one(i, p);
      } else {
        if (z[i].count(p)) continue;
        zero(i, p);
      }
    }
    for (int i = 0; i < 20; i++) ans += sum[i] * (1 << i);
    cout << ans << "\n";
  }
  return 0;
}