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;
struct ST {
  int l, r;
  int len;
  long long sum;
} tr[100005 << 2];
int n, m;
int a[100005], p[100005], v[100005];
void push_up(int l, int r, int rt) {
  tr[rt].sum = tr[rt << 1].sum + tr[rt << 1 | 1].sum;
  long long L = tr[rt << 1].r;
  long long R = tr[rt << 1 | 1].l;
  tr[rt].sum -= L * (L + 1) / 2;
  tr[rt].sum -= R * (R + 1) / 2;
  tr[rt].sum += (L + R) * (L + R + 1) / 2;
  int mid = (l + r) >> 1;
  if (tr[rt << 1].r == (mid - l + 1)) {
    tr[rt].l = tr[rt << 1].r + tr[rt << 1 | 1].l;
  } else
    tr[rt].l = tr[rt << 1].l;
  if (tr[rt << 1 | 1].l == (r - mid)) {
    tr[rt].r = tr[rt << 1 | 1].l + tr[rt << 1].r;
  } else {
    tr[rt].r = tr[rt << 1 | 1].r;
  }
}
void build(int l, int r, int rt, int i) {
  if (l == r) {
    tr[rt].sum = tr[rt].l = tr[rt].r = ((a[l] >> i) & 1);
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1, i);
  build(mid + 1, r, rt << 1 | 1, i);
  push_up(l, r, rt);
}
void update(int l, int r, int rt, int pos, int x) {
  if (l == r) {
    tr[rt].l = tr[rt].r = tr[rt].sum = x;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    update(l, mid, rt << 1, pos, x);
  else
    update(mid + 1, r, rt << 1 | 1, pos, x);
  push_up(l, r, rt);
}
long long cnt[18][100005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &p[i], &v[i]);
  }
  for (int i = 0; i < 18; i++) {
    build(1, n, 1, i);
    for (int j = 1; j <= m; j++) {
      update(1, n, 1, p[j], (v[j] >> i) & 1);
      cnt[i][j] = tr[1].sum;
    }
  }
  for (int i = 1; i <= m; i++) {
    long long ans = 0;
    for (int j = 0; j < 18; j++) {
      ans += cnt[j][i] * (1 << j);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}