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 maxn = 17, maxd = 1e5 + 10;
int n, m, a[maxd], ipt[maxd];
struct SegTree {
  long long sum[maxd << 2], lmax[maxd << 2], rmax[maxd << 2];
  void build(int l, int r, int rt) {
    if (l == r) {
      sum[rt] = lmax[rt] = rmax[rt] = a[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    pushUp(l, r, mid, rt);
    return;
  }
  void pushUp(int l, int r, int m, int rt) {
    sum[rt] =
        sum[rt << 1] + sum[rt << 1 | 1] + rmax[rt << 1] * lmax[rt << 1 | 1];
    lmax[rt] = lmax[rt << 1];
    if (lmax[rt] == m - l + 1) {
      lmax[rt] += lmax[rt << 1 | 1];
    }
    rmax[rt] = rmax[rt << 1 | 1];
    if (rmax[rt] == r - m) {
      rmax[rt] += rmax[rt << 1];
    }
  }
  void update(int x, int d, int l, int r, int rt) {
    if (l == r) {
      sum[rt] = lmax[rt] = rmax[rt] = d;
      return;
    }
    int mid = (l + r) >> 1;
    if (x <= mid)
      update(x, d, l, mid, rt << 1);
    else
      update(x, d, mid + 1, r, rt << 1 | 1);
    pushUp(l, r, mid, rt);
    return;
  }
} segtree[maxn];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &ipt[i]);
  }
  for (int i = 0; i < maxn; ++i) {
    for (int j = 1; j <= n; ++j) {
      a[j] = (ipt[j] >> i) & 1;
    }
    segtree[i].build(1, n, 1);
  }
  int x, d;
  long long ans = 0;
  while (m--) {
    ans = 0;
    scanf("%d%d", &x, &d);
    for (int i = 0; i < maxn; ++i) {
      segtree[i].update(x, (d >> i) & 1, 1, n, 1);
      ans += segtree[i].sum[1] << i;
    }
    printf("%I64d\n", ans);
  }
}