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 MOD = 1e9 + 7;
long long N, Q;
long long pos[200005], W[200005];
struct segtree {
  long long sum[800005], sumL[800005], sumR[800005];
  long long upd_pos, ql, qr, type;
  long long val, res;
  void build(long long n, long long l, long long r) {
    if (l == r) {
      sum[n] = W[l];
      sumL[n] = W[l] * (pos[l] + (N - l + 1)) % MOD;
      sumR[n] = W[l] * (pos[l] - l) % MOD;
      return;
    }
    long long mid = (l + r) >> 1;
    build((n << 1), l, mid);
    build((n << 1 | 1), mid + 1, r);
    sum[n] = sum[(n << 1)] + sum[(n << 1 | 1)];
    sumL[n] = sumL[(n << 1)] + sumL[(n << 1 | 1)] % MOD;
    sumR[n] = sumR[(n << 1)] + sumR[(n << 1 | 1)] % MOD;
  }
  void build() { build(1, 1, N); }
  void update(long long n, long long l, long long r) {
    if (l == r) {
      sum[n] = val;
      sumL[n] = val * (pos[l] + (N - l + 1)) % MOD;
      sumR[n] = val * (pos[l] - l) % MOD;
      return;
    }
    long long mid = (l + r) >> 1;
    if (upd_pos <= mid)
      update((n << 1), l, mid);
    else
      update((n << 1 | 1), mid + 1, r);
    sum[n] = sum[(n << 1)] + sum[(n << 1 | 1)];
    sumL[n] = sumL[(n << 1)] + sumL[(n << 1 | 1)] % MOD;
    sumR[n] = sumR[(n << 1)] + sumR[(n << 1 | 1)] % MOD;
  }
  void update(long long pos, long long new_val) {
    upd_pos = pos;
    val = new_val;
    update(1, 1, N);
  }
  void get_sum(long long n, long long l, long long r) {
    if (ql <= l && r <= qr) {
      res += (type == 1 ? sum[n] : type == 2 ? sumL[n] : sumR[n]);
      return;
    }
    long long mid = (l + r) >> 1;
    if (ql <= mid) get_sum((n << 1), l, mid);
    if (qr > mid) get_sum((n << 1 | 1), mid + 1, r);
  }
  long long query(long long l, long long r, long long query_type) {
    if (l > r) return 0;
    ql = l, qr = r;
    type = query_type;
    res = 0;
    get_sum(1, 1, N);
    if (type == 1)
      return res;
    else
      return res % MOD;
  }
  long long K;
  void binser(long long n, long long l, long long r) {
    if (K < 0) return;
    if (l == r) {
      if (K < sum[n]) {
        res = l;
        K = -1;
      } else {
        K -= sum[n];
      }
      return;
    }
    long long mid = (l + r) >> 1;
    if (ql <= l && r <= qr) {
      if (sum[n] <= K) {
        K -= sum[n];
        return;
      }
      if (K < sum[(n << 1)]) {
        binser((n << 1), l, mid);
      } else {
        K -= sum[(n << 1)];
        binser((n << 1 | 1), mid + 1, r);
      }
      return;
    }
    if (ql <= mid) binser((n << 1), l, mid);
    if (qr > mid) binser((n << 1 | 1), mid + 1, r);
  }
  long long cari(long long l, long long r, long long sum) {
    K = sum >> 1;
    ql = l, qr = r;
    binser(1, 1, N);
    return res;
  }
};
segtree tree;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> Q;
  for (long long i = 1; i <= N; i++) {
    cin >> pos[i];
  }
  for (long long i = 1; i <= N; i++) {
    cin >> W[i];
  }
  tree.build();
  long long x, y, mid;
  long long total_sum, sum_left, sum_right;
  while (Q--) {
    cin >> x >> y;
    if (x < 0) {
      tree.update(-x, y);
    } else {
      total_sum = tree.query(x, y, 1);
      mid = tree.cari(x, y, total_sum);
      sum_left =
          tree.query(x, mid - 1, 1) % MOD * (pos[mid] + (N - mid + 1)) % MOD -
          tree.query(x, mid - 1, 2);
      sum_right = tree.query(mid + 1, y, 3) -
                  tree.query(mid + 1, y, 1) % MOD * (pos[mid] - mid) % MOD;
      while (sum_left < 0) sum_left = sum_left + MOD;
      while (sum_right < 0) sum_right = sum_right + MOD;
      cout << (sum_left + sum_right) % MOD << "\n";
    }
  }
  cin >> N;
}