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;
template <class T>
T sqr(T x) {
  return x * x;
}
const int maxn = 101001;
const int maxm = 21;
long long two[31];
long long a[maxn];
int n, x, y, m;
inline long long gao(long long x) { return x * (x + 1) / 2; }
int main() {
  two[0] = 1;
  for (int i = 1; i <= 20; i++) two[i] = two[i - 1] * 2;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
  }
  long long ans = 0;
  for (int k = 0; k <= 18; k++)
    for (int i = 1; i <= n; i++)
      if ((a[i] >> k) & 1) {
        for (int j = i; j <= n; j++) {
          if (((a[j + 1] >> k) & 1) == 0) {
            ans += gao(j - i + 1) * two[k];
            i = j;
            break;
          }
        }
      }
  for (int ii = 1; ii <= m; ii++) {
    scanf("%d%d", &x, &y);
    for (int k = 0; k <= 18; k++)
      if (((y >> k) & 1) != ((a[x] >> k) & 1)) {
        int cnt1 = 0, cnt2 = 0;
        for (int i = x - 1; i >= 1 && ((a[i] >> k) & 1); i--, cnt1++)
          ;
        for (int i = x + 1; i <= n && ((a[i] >> k) & 1); i++, cnt2++)
          ;
        if (((y >> k) & 1) == 0)
          ans = ans + (-gao(cnt1 + cnt2 + 1) + gao(cnt1) + gao(cnt2)) * two[k];
        else
          ans = ans + (gao(cnt1 + cnt2 + 1) - gao(cnt1) - gao(cnt2)) * two[k];
      }
    a[x] = y;
    printf("%I64d\n", ans);
  }
  fclose(stdin);
  return 0;
}