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 N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);
template <class T, class S>
inline void add(T& a, S b) {
  a += b;
  if (a >= mod) a -= mod;
}
template <class T, class S>
inline void sub(T& a, S b) {
  a -= b;
  if (a < 0) a += mod;
}
template <class T, class S>
inline bool chkmax(T& a, S b) {
  return a < b ? a = b, true : false;
}
template <class T, class S>
inline bool chkmin(T& a, S b) {
  return a > b ? a = b, true : false;
}
int n, q, a[N], w[N];
struct Bit {
  long long a[N];
  void modify(int x, int v) {
    for (int i = x; i <= n; i += i & -i) a[i] += v;
  }
  long long sum(int x) {
    long long ans = 0;
    for (int i = x; i; i -= i & -i) ans += a[i];
    return ans;
  }
  long long query(int L, int R) {
    if (L > R) return 0;
    return sum(R) - sum(L - 1);
  }
} bit;
struct Bit2 {
  int a[N];
  void modify(int x, int v) {
    for (int i = x; i <= n; i += i & -i) add(a[i], v);
  }
  int sum(int x) {
    long long ans = 0;
    for (int i = x; i; i -= i & -i) add(ans, a[i]);
    return ans;
  }
  int query(int L, int R) {
    if (L > R) return 0;
    return (sum(R) - sum(L - 1) + mod) % mod;
  }
} prefix, suffix;
int main() {
  long long ret = 0;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  for (int i = 1; i <= n; i++) {
    bit.modify(i, w[i]);
    prefix.modify(i, 1LL * abs(a[i] - a[1] - i + 1) * w[i] % mod);
  }
  for (int i = n; i >= 1; i--)
    suffix.modify(i, 1LL * abs(a[n] - a[i] - n + i) * w[i] % mod);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x = -x;
      bit.modify(x, -w[x]);
      prefix.modify(x, mod - (1LL * abs(a[x] - a[1] - x + 1) * w[x] % mod));
      suffix.modify(x, mod - (1LL * abs(a[n] - a[x] - n + x) * w[x] % mod));
      w[x] = y;
      bit.modify(x, w[x]);
      prefix.modify(x, 1LL * abs(a[x] - a[1] - x + 1) * w[x] % mod);
      suffix.modify(x, 1LL * abs(a[n] - a[x] - n + x) * w[x] % mod);
    } else {
      int len = y - x + 1;
      int low = a[x], high = a[y], p = low, low2, high2, p2, fen;
      while (low <= high) {
        int mid = low + high >> 1;
        low2 = x, high2 = y, p2 = y + 1;
        while (low2 <= high2) {
          int mid2 = low2 + high2 >> 1;
          if (a[mid2] > mid + mid2 - x)
            p2 = mid2, high2 = mid2 - 1;
          else
            low2 = mid2 + 1;
        }
        long long valL = bit.query(x, p2 - 1);
        long long valR = bit.query(p2, y);
        if (valL > valR)
          p = mid, high = mid - 1, fen = p2;
        else
          low = mid + 1;
      }
      long long ans = 0;
      if (fen <= y) {
        add(ans, prefix.query(fen, y));
        sub(ans, bit.query(fen, y) % mod * (p - (a[1] + x - 1)) % mod);
      }
      if (fen - 1 >= x) {
        add(ans, suffix.query(x, fen - 1));
        sub(ans, bit.query(x, fen - 1) % mod * (a[n] - (n - x) - p) % mod);
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}