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 + 5, M = 18;
int n, m, a[N], x, y;
struct item {
  long long sum;
  int pref, suf;
  bool full;
  item(int x = 0) { sum = pref = suf = full = x; }
};
item t[M][4 * N];
inline bool bit(int x, int y) { return x & (1ll << y); }
inline item mrg(item a, item b) {
  item cur;
  cur.sum = a.sum + b.sum + 1ll * a.suf * b.pref;
  cur.pref = a.pref;
  cur.suf = b.suf;
  if (a.full) {
    cur.pref += b.pref;
  }
  if (b.full) {
    cur.suf += a.suf;
  }
  cur.full = a.full & b.full;
  return cur;
}
void build(int x, int v, int l, int r) {
  if (l == r) {
    t[x][v] = item(0);
    return;
  }
  int mid = (r + l) >> 1;
  build(x, v + v, l, mid);
  build(x, v + v + 1, mid + 1, r);
  t[x][v] = mrg(t[x][v + v], t[x][v + v + 1]);
}
void update(int x, int v, int l, int r, int pos, int val) {
  if (l == r) {
    t[x][v] = item(val);
    return;
  }
  int mid = (r + l) >> 1;
  if (pos <= mid) {
    update(x, v + v, l, mid, pos, val);
  } else {
    update(x, v + v + 1, mid + 1, r, pos, val);
  }
  t[x][v] = mrg(t[x][v + v], t[x][v + v + 1]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < M; i++) {
    build(i, 1, 1, n);
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    for (int j = 0; j < M; j++) {
      update(j, 1, 1, n, i, bit(a[i], j));
    }
  }
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    a[x] = y;
    long long ans = 0;
    for (int j = 0; j < M; j++) {
      update(j, 1, 1, n, x, bit(a[x], j));
      ans += (t[j][1].sum << j);
    }
    cout << ans << "\n";
  }
}