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;
constexpr int B = 20, N = 1e5 + 7;
int cnt = 0, n, Q;
pair<int, int> l[B][4 * N], r[B][4 * N];
inline pair<pair<int, int>, pair<int, int> > query(int i, int pos, int tag = 1,
                                                   int s = 0, int e = n + 2) {
  if (e - s == 1) return make_pair(l[i][tag], r[i][tag]);
  int mid = (s + e) / 2;
  if (pos < mid) {
    pair<pair<int, int>, pair<int, int> > p = query(i, pos, tag * 2, s, mid);
    p.first = max(p.first, l[i][tag]);
    p.second = max(p.second, r[i][tag]);
    return p;
  } else {
    pair<pair<int, int>, pair<int, int> > p =
        query(i, pos, tag * 2 + 1, mid, e);
    p.first = max(p.first, l[i][tag]);
    p.second = max(p.second, r[i][tag]);
    return p;
  }
}
inline void update(int i, int ll, int rr, char c, int nw, int tag = 1,
                   int s = 0, int e = n + 2) {
  if (tag == 1) cnt++;
  if (s >= rr || e <= ll) return;
  if (s >= ll && e <= rr) {
    if (c == 'l')
      l[i][tag] = make_pair(cnt, nw);
    else
      r[i][tag] = make_pair(cnt, nw);
    return;
  }
  int mid = (s + e) / 2;
  update(i, ll, rr, c, nw, tag * 2, s, mid);
  update(i, ll, rr, c, nw, tag * 2 + 1, mid, e);
}
int a[N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin >> n >> Q;
  int last[B] = {};
  a[n + 1] = a[0] = 0;
  long long ans = 0;
  for (int i = 1; i <= n + 1; i++) {
    if (i <= n) cin >> a[i];
    for (int j = 0; j < B; j++)
      if (!(a[i] & (1 << j))) {
        update(j, last[j], i, 'r', i);
        for (int x = last[j] + 1; x < i; x++) ans += ((1 << j) * 1ll * (i - x));
        last[j] = i;
      }
  }
  for (int i = 0; i < B; i++) last[i] = n + 1;
  for (int i = n; i >= 0; --i) {
    for (int j = 0; j < B; j++)
      if (!(a[i] & (1 << j))) {
        update(j, i, last[j] + 1, 'l', i);
        last[j] = i;
      }
  }
  while (Q--) {
    int x, y;
    cin >> x >> y;
    for (int i = 0; i < B; i++) {
      if ((a[x] & (1 << i)) && !(y & (1 << i))) {
        pair<pair<int, int>, pair<int, int> > oo = query(i, x);
        pair<int, int> q = {oo.first.second, oo.second.second};
        ans -= ((1 << i) * 1ll * (x - q.first) * (q.second - x));
        update(i, q.first, x, 'r', x);
        update(i, x + 1, q.second + 1, 'l', x);
      }
      if (!(a[x] & (1 << i)) && (y & (1 << i))) {
        pair<pair<int, int>, pair<int, int> > oo = query(i, x);
        pair<int, int> q = {oo.first.second, oo.second.second};
        ans += ((1 << i) * 1ll * (x - q.first) * (q.second - x));
        update(i, q.first, q.second, 'r', q.second);
        update(i, q.first + 1, q.second + 1, 'l', q.first);
      }
    }
    a[x] = y;
    cout << ans << '\n';
  }
}