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 = 0x3f3f3f3f, mod = 1000000007;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const double pi = acosl(-1.), eps = 1e-9;
inline int power(int a, int b, int m = mod, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
const int NN = 202020;
long long a[NN], w[NN];
int aw[NN];
int sw[NN];
long long W[NN];
void upaw(int u, int d) {
  if (d < 0) d += mod;
  for (; u < NN; u += u & -u) {
    aw[u] += d;
    if (aw[u] >= mod) aw[u] -= mod;
  }
}
void upsw(int u, int d) {
  if (d < 0) d += mod;
  for (; u < NN; u += u & -u) {
    sw[u] += d;
    if (sw[u] >= mod) sw[u] -= mod;
  }
}
void upW(int u, long long d) {
  for (; u < NN; u += u & -u) W[u] += d;
}
int caaw(int u, int ans = 0) {
  for (; u; u -= u & -u) {
    ans += aw[u];
    if (ans >= mod) ans -= mod;
  }
  return ans;
}
int casw(int u, int ans = 0) {
  for (; u; u -= u & -u) {
    ans += sw[u];
    if (ans >= mod) ans -= mod;
  }
  return ans;
}
long long caW(int u, long long ans = 0) {
  for (; u; u -= u & -u) ans += W[u];
  return ans;
}
int nam(int L, int R, int id) {
  long long lft = caW(id - 1) - caW(L - 1);
  long long rgt = caW(R) - caW(id);
  if (lft == rgt) return 0;
  if (lft < rgt) return -1;
  if (lft > rgt) return 1;
}
inline void sum(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
inline void sub(int &a, int b) {
  a += b;
  if (a < 0) a += mod;
}
int calc(int L, int R, int id) {
  long long ans = 0;
  ans += caaw(R) - caaw(id - 1);
  ans += (id - a[id]) * ((caW(R) - caW(id - 1)) % mod) % mod;
  ans -= casw(R) - casw(id - 1);
  ans += (a[id] - id) * ((caW(id) - caW(L - 1)) % mod) % mod;
  ans -= caaw(id) - caaw(L - 1);
  ans += casw(id) - casw(L - 1);
  ans %= mod;
  if (ans < 0) ans += mod;
  return ans;
}
int main() {
  int n, T;
  cin >> n >> T;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) {
    scanf("%d", w + i);
    upaw(i, (int)((long long)a[i] * w[i] % mod));
    upsw(i, (int)((long long)i * w[i] % mod));
    upW(i, w[i]);
  }
  int L, R;
  while (T--) {
    scanf("%d%d", &L, &R);
    if (L < 0) {
      L = -L;
      upaw(L, (int)(((long long)a[L] * R - (long long)a[L] * w[L]) % mod));
      upsw(L, ((long long)L * R - (long long)L * w[L]) % mod);
      upW(L, R - w[L]);
      w[L] = R;
      continue;
    }
    int st = L, ed = R + 1, md1, md2;
    int d, d0;
    while (st + 2 < ed) {
      md1 = ((st << 1) + ed) / 3;
      md2 = (st + (ed << 1)) / 3;
      d = nam(L, R, md1);
      if (d < 0) {
        st = md1;
        continue;
      }
      if (d > 0) {
        ed = md2;
        continue;
      }
      d = nam(L, R, md2);
      if (d < 0) {
        st = md1;
        continue;
      }
      if (d > 0) {
        ed = md2;
        continue;
      }
      ed = md2;
    }
    long long lft = caW(st - 1) - caW(L - 1);
    long long rgt = caW(R) - caW(st);
    if (lft + w[st] - rgt < 0) st++;
    printf("%d\n", calc(L, R, st));
  }
  return 0;
}