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;
using ll = long long;
using ld = long double;
constexpr ll mod = 1000000007;
struct modft {
  vector<ll> tree;
  void update(int i, ll val) {
    i++;
    while (i < (int)(tree).size()) {
      tree[i] += val;
      tree[i] %= mod;
      i += (i & (-i));
    }
  }
  modft(int n) { tree.assign(n + 1, 0); }
  ll prefix_sum(int i) {
    if (i >= (int)(tree).size()) i = (int)(tree).size() - 1;
    ll sum = 0;
    i++;
    while (i > 0) {
      sum += tree[i];
      sum %= mod;
      i -= (i & (-i));
    }
    return sum;
  }
  ll query(int l, int r) {
    --r;
    --l;
    ll res = prefix_sum(r);
    res -= prefix_sum(l);
    return ((res % mod) + mod) % mod;
  }
  void set(int i, ll val) {
    ll cur = query(i, i + 1);
    update(i, (((val - cur) % mod) + mod) % mod);
  }
};
struct ft {
  vector<ll> tree;
  void update(int i, ll val) {
    i++;
    while (i < (int)(tree).size()) {
      tree[i] += val;
      i += (i & (-i));
    }
  }
  ft(int n) { tree.assign(n + 1, 0); }
  ll prefix_sum(int i) {
    if (i >= (int)(tree).size()) i = (int)(tree).size() - 1;
    ll sum = 0;
    i++;
    while (i > 0) {
      sum += tree[i];
      i -= (i & (-i));
    }
    return sum;
  }
  ll query(int l, int r) {
    --r;
    --l;
    ll res = prefix_sum(r);
    res -= prefix_sum(l);
    return res;
  }
  void set(int i, ll val) {
    ll cur = query(i, i + 1);
    update(i, val - cur);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  vector<ll> a(n), w(n);
  ft raw(n);
  modft lin(n), space(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    cin >> w[i];
    raw.update(i, w[i]);
    lin.update(i, (i * w[i]) % mod);
    space.update(i, (a[i] * w[i]) % mod);
  }
  for (int _ = 0; _ < q; _++) {
    ll x, y;
    cin >> x >> y;
    if (x < 0) {
      int i = (int)(-x - 1);
      w[i] = y;
      raw.set(i, w[i]);
      lin.set(i, (i * w[i]) % mod);
      space.set(i, (a[i] * w[i]) % mod);
    } else {
      --x;
      int lo = (int)x;
      int hi = (int)y;
      ll sum = raw.query(lo, hi);
      while (hi - lo > 1) {
        int mid = (lo + hi) / 2;
        if (raw.query((int)x, mid) > sum / 2) {
          hi = mid;
        } else {
          lo = mid;
        }
      }
      int _x = (int)x;
      int _y = (int)y;
      int k = lo;
      if (raw.query(_x, k + 1) <= sum / 2) cout << "?" << endl;
      if (raw.query(_x, k) > sum / 2) cout << "?" << endl;
      ll left = (((a[k] - k) % mod) * (raw.query(_x, k) % mod)) % mod;
      left = (left - space.query(_x, k)) % mod;
      left = (left + lin.query(_x, k)) % mod;
      ll right = (((k - a[k]) % mod) * (raw.query(k + 1, _y) % mod)) % mod;
      right = (right + space.query(k + 1, _y)) % mod;
      right = (right - lin.query(k + 1, _y)) % mod;
      ll ans = (((left + right) % mod) + mod) % mod;
      cout << ans << '\n';
    }
  }
}