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 + 10, mod = 1e9 + 7;
int n, q, tmp[N], a[N], b[N], w[N], val[N];
int lowbit(int x) { return x & -x; }
struct BITs {
  long long c[N];
  void ins(int x, int v) {
    for (; x <= n; x += lowbit(x)) c[x] += v;
  }
  long long query(int x) {
    long long ret = 0;
    for (; x; x -= lowbit(x)) ret += c[x];
    return ret;
  }
  long long calc(int l, int r) { return query(r) - query(l - 1); }
} tr1;
struct BITssum {
  int c[N];
  void ins(int x, int v) {
    for (; x <= n; x += lowbit(x)) c[x] = (c[x] + v) % mod;
  }
  int query(int x) {
    int ret = 0;
    for (; x; x -= lowbit(x)) ret = (ret + c[x]) % mod;
    return ret;
  }
  int calc(int l, int r) { return (query(r) - query(l - 1)) % mod; }
} tr2;
bool check(int x, int l, int r, int mid, long long sum) {
  if (tr1.query(mid) - tr1.query(l - 1) >= (sum + 1) / 2) return 1;
  return 0;
}
int solve(int L, int R) {
  long long cnt = tr1.calc(L, R);
  int l = 1, r = n, best = n;
  while (l <= r) {
    int mid = l + r >> 1;
    if (check(mid, L, R, mid, cnt))
      best = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  int ans = 0;
  ans = (ans + (1ll * tr1.calc(L, best - 1) % mod * val[best] % mod -
                tr2.calc(L, best - 1)) %
                   mod) %
        mod;
  ans = (ans + (tr2.calc(best + 1, R) -
                1ll * tr1.calc(best + 1, R) % mod * val[best] % mod) %
                   mod) %
        mod;
  return (ans + mod) % mod;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), tmp[i] = a[i];
  sort(tmp + 1, tmp + n + 1);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  for (int i = 1; i <= n; i++) {
    b[i] = lower_bound(tmp + 1, tmp + n + 1, a[i]) - tmp, val[i] = a[i] - i;
    tr1.ins(b[i], w[i]);
    tr2.ins(b[i], 1ll * w[i] * val[i] % mod);
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x = -x;
      tr1.ins(b[x], -w[x]), tr2.ins(b[x], -1ll * w[x] * val[x] % mod);
      w[x] = y;
      tr1.ins(b[x], w[x]), tr2.ins(b[x], 1ll * w[x] * val[x] % mod);
    } else
      printf("%d\n", solve(x, y));
  }
  return 0;
}