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 long long int N = 200009, mod = 1e9 + 7;
int n, q, a[N], low[4 * N], high[4 * N], leaf[N];
long long int b[N], w[N], aw[N], sw[4 * N], saw[4 * N], sb[4 * N];
void Built(int v, int l, int h) {
  low[v] = l;
  high[v] = h;
  if (l == h) {
    sw[v] = w[l];
    saw[v] = aw[l];
    leaf[l] = v;
    sb[v] = b[l];
  } else {
    long long int m = (l + h) / 2;
    Built(v * 2, l, m);
    Built(v * 2 + 1, m + 1, h);
    sw[v] = (sw[v * 2] + sw[v * 2 + 1]);
    saw[v] = (saw[v * 2] + saw[v * 2 + 1]);
    sb[v] = (sb[v * 2] + sb[v * 2 + 1]);
  }
}
long long int Next() {
  register char ch = getchar();
  bool neg = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      neg = 1;
    }
    ch = getchar();
  }
  long long int l = 0;
  while (ch >= '0' && ch <= '9') {
    l = l * 10 + ch - '0';
    ch = getchar();
  }
  if (neg) {
    return -l;
  }
  return l;
}
void Inp() {
  n = Next();
  q = Next();
  for (int i = 1; i <= n; i++) {
    a[i] = Next();
  }
  for (int i = 1; i <= n; i++) {
    w[i] = Next();
    aw[i] = (w[i] * i) % mod;
    b[i] = (w[i] * a[i]) % mod;
  }
  Built(1, 1, n);
}
long long int Update(int k) {
  long long int v = leaf[k];
  sw[v] = w[k];
  saw[v] = aw[k];
  sb[v] = b[k];
  v = v / 2;
  while (v != 0) {
    sw[v] = (sw[v * 2] + sw[v * 2 + 1]);
    saw[v] = (saw[v * 2] + saw[v * 2 + 1]);
    sb[v] = (sb[v * 2] + sb[v * 2 + 1]);
    v = v / 2;
  }
}
long long int Sumw(int v, int l, int h) {
  if (low[v] > h || high[v] < l) {
    return 0;
  }
  if (low[v] >= l && high[v] <= h) {
    return sw[v];
  }
  return Sumw(v * 2, l, h) + Sumw(v * 2 + 1, l, h);
}
long long int Sumb(int v, int l, int h) {
  if (low[v] > h || high[v] < l) {
    return 0;
  }
  if (low[v] >= l && high[v] <= h) {
    return sb[v];
  }
  return (Sumb(v * 2, l, h) + Sumb(v * 2 + 1, l, h));
}
long long int Sumaw(int v, int l, int h) {
  if (low[v] > h || high[v] < l) {
    return 0;
  }
  if (low[v] >= l && high[v] <= h) {
    return saw[v];
  }
  return (Sumaw(v * 2, l, h) + Sumaw(v * 2 + 1, l, h));
}
long long int Tinhaw(int l, int r, int k) {
  return (Sumaw(1, l, r) + (k - l) * Sumw(1, l, r));
}
long long int Calc1(int l, int r, int k) {
  long long int cnt = (r - l + 1);
  return ((Sumw(1, l, r) % mod) * k - Sumb(1, l, r) + Tinhaw(l, r, 0));
}
long long int Calc2(int l, int r, int k, int ol) {
  if (r < l) {
    return 0;
  }
  return (-Sumb(1, l, r) + (Sumw(1, l, r) % mod) * k + Tinhaw(l, r, l - ol));
}
long long int Calc3(int l, int r, int k, int ol) {
  return (Sumb(1, l, r) - (Sumw(1, l, r) % mod) * k - Tinhaw(l, r, l - ol));
}
long long int Calc(int l, int r, int k) {
  int v = upper_bound(a + l, a + r + 1, k) - a - 1, low = v + 1, high = r;
  while (low <= high) {
    long long int mid = (low + high) / 2;
    if (a[mid] <= k + mid - l) {
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  return (Calc1(l, v, k) + Calc2(v + 1, high, k, l) +
          Calc3(high + 1, r, k, l)) %
         mod;
}
void Query(int p, int q) {
  int l = p, h = q;
  long long int s = Sumw(1, p, q);
  while (l <= h) {
    long long int m = (l + h) / 2;
    if (Sumw(1, p, m) <= s / 2) {
      l = m + 1;
    } else {
      h = m - 1;
    }
  }
  cout << (Calc(p, q, a[l] - (l - p)) + mod) % mod << '\n';
}
void Solve() {
  for (int i = 1; i <= q; i++) {
    int x, y;
    x = Next();
    y = Next();
    if (x < 0) {
      x = -x;
      w[x] = y;
      b[x] = (a[x] * w[x]) % mod;
      aw[x] = (w[x] * x) % mod;
      Update(x);
    } else {
      Query(x, y);
    }
  }
}
int main() {
  Inp();
  Solve();
}