Problem: There is an infinite line consisting of cells. There are n boxes in some cells of this line. The i-th box stands in the cell a_i and has weight w_i. All a_i are distinct, moreover, a_{i - 1} < a_i holds for all valid i.

You would like to put together some boxes. Putting together boxes with indices in the segment [l, r] means that you will move some of them in such a way that their positions will form some segment [x, x + (r - l)].

In one step you can move any box to a neighboring cell if it isn't occupied by another box (i.e. you can choose i and change a_i by 1, all positions should remain distinct). You spend w_i units of energy moving the box i by one cell. You can move any box any number of times, in arbitrary order.

Sometimes weights of some boxes change, so you have queries of two types: 

  1. id nw — weight w_{id} of the box id becomes nw. 
  2. l r — you should compute the minimum total energy needed to put together boxes with indices in [l, r]. Since the answer can be rather big, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7. Note that the boxes are not moved during the query, you only should compute the answer. 



Note that you should minimize the answer, not its remainder modulo 10^9 + 7. So if you have two possible answers 2 ⋅ 10^9 + 13 and 2 ⋅ 10^9 + 14, you should choose the first one and print 10^9 + 6, even though the remainder of the second answer is 0.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the number of boxes and the number of queries.

The second line contains n integers a_1, a_2, ... a_n (1 ≤ a_i ≤ 10^9) — the positions of the boxes. All a_i are distinct, a_{i - 1} < a_i holds for all valid i.

The third line contains n integers w_1, w_2, ... w_n (1 ≤ w_i ≤ 10^9) — the initial weights of the boxes.

Next q lines describe queries, one query per line.

Each query is described in a single line, containing two integers x and y. If x < 0, then this query is of the first type, where id = -x, nw = y (1 ≤ id ≤ n, 1 ≤ nw ≤ 10^9). If x > 0, then the query is of the second type, where l = x and r = y (1 ≤ l_j ≤ r_j ≤ n). x can not be equal to 0.

Output

For each query of the second type print the answer on a separate line. Since answer can be large, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7.

Example

Input

5 8
1 2 6 7 10
1 1 1 1 2
1 1
1 5
1 3
3 5
-3 5
-1 10
1 4
2 5


Output

0
10
3
4
18
7

Note

Let's go through queries of the example: 

  1. 1\ 1 — there is only one box so we don't need to move anything. 
  2. 1\ 5 — we can move boxes to segment [4, 8]: 1 ⋅ |1 - 4| + 1 ⋅ |2 - 5| + 1 ⋅ |6 - 6| + 1 ⋅ |7 - 7| + 2 ⋅ |10 - 8| = 10. 
  3. 1\ 3 — we can move boxes to segment [1, 3]. 
  4. 3\ 5 — we can move boxes to segment [7, 9]. 
  5. -3\ 5 — w_3 is changed from 1 to 5. 
  6. -1\ 10 — w_1 is changed from 1 to 10. The weights are now equal to w = [10, 1, 5, 1, 2]. 
  7. 1\ 4 — we can move boxes to segment [1, 4]. 
  8. 2\ 5 — we can move boxes to segment [5, 8]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
long long read_long() {
  long long ans = 0;
  char c;
  bool started = false;
  bool is_negative = false;
  while (true) {
    c = getchar();
    if ((c < '0' || c > '9') && c != '-' && !started) continue;
    if ((c < '0' || c > '9') && c != '-' && started) break;
    if (started) {
      ans = ans * 10;
    }
    started = true;
    if (c == '-')
      is_negative = true;
    else
      ans += c - '0';
  }
  if (is_negative) ans *= -1;
  return ans;
}
int n, q;
int a[200200];
int b[200200];
int w[200200];
long long tw[200200 * 4];
int t[200200 * 4];
void build_tw(int v, int tl, int tr) {
  if (tl == tr) {
    tw[v] = w[tl];
    t[v] = w[tl] * 1ll * b[tl] % mod;
  } else {
    int mid = (tl + tr) >> 1;
    build_tw(2 * v, tl, mid);
    build_tw(2 * v + 1, mid + 1, tr);
    tw[v] = tw[v * 2] + tw[v * 2 + 1];
    t[v] = t[2 * v] + t[2 * v + 1];
    t[v] %= mod;
  }
}
long long get_tw(int v, int tl, int tr, int l, int r) {
  if (l > tr || r < tl) return 0;
  if (l <= tl && tr <= r) return tw[v];
  int mid = (tl + tr) >> 1;
  long long ans1 = get_tw(2 * v, tl, mid, l, r);
  long long ans2 = get_tw(2 * v + 1, mid + 1, tr, l, r);
  return ans1 + ans2;
}
int get(int v, int tl, int tr, int l, int r) {
  if (l > tr || r < tl) return 0;
  if (l <= tl && tr <= r) return t[v];
  int mid = (tl + tr) >> 1;
  int ans1 = get(2 * v, tl, mid, l, r);
  int ans2 = get(2 * v + 1, mid + 1, tr, l, r);
  return (ans1 + ans2) % mod;
}
void update_tw(int v, int tl, int tr, int pos, int val) {
  if (tl == tr) {
    tw[v] = val;
    t[v] = val * 1ll * b[tl] % mod;
  } else {
    int mid = (tl + tr) >> 1;
    if (pos <= mid)
      update_tw(2 * v, tl, mid, pos, val);
    else
      update_tw(2 * v + 1, mid + 1, tr, pos, val);
    tw[v] = tw[v * 2] + tw[v * 2 + 1];
    t[v] = t[2 * v] + t[2 * v + 1];
    t[v] %= mod;
  }
}
void precalc() { build_tw(1, 0, n - 1); }
void update(int pos, int val) { update_tw(1, 0, n - 1, pos, val); }
int get_mid(int l, int r) {
  long long S = get_tw(1, 0, n - 1, l, r);
  int L = l, R = r, ans = r;
  while (L <= R) {
    int mid = (L + R) >> 1;
    long long cur = get_tw(1, 0, n - 1, l, mid);
    if (cur * 2 >= S) {
      ans = mid;
      R = mid - 1;
    } else
      L = mid + 1;
  }
  return ans;
}
int get(int l, int r) {
  if (l >= r) return 0;
  int mid = get_mid(l, r);
  int L = get_tw(1, 0, n - 1, l, mid - 1) % mod;
  int R = get_tw(1, 0, n - 1, mid + 1, r) % mod;
  int Lw = get(1, 0, n - 1, l, mid - 1);
  int Rw = get(1, 0, n - 1, mid + 1, r);
  int ans = b[mid] * 1ll * L % mod;
  ans -= Lw;
  ans %= mod;
  ans += Rw;
  ans %= mod;
  ans -= b[mid] * 1ll * R % mod;
  ans %= mod;
  if (ans < 0) ans += mod;
  return ans;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    b[i] = a[i] - i;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", w + i);
  }
  precalc();
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      update(-x - 1, y);
    } else {
      printf("%d\n", get(x - 1, y - 1));
    }
  }
  return 0;
}