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 = 1000000000 + 7;
const long long LL_INF = 1ll * INF * INF;
const int MAX_N = 300000 + 7;
template <typename T>
inline void addmod(T& a, const long long& b, const int& MOD = INF) {
  a = (a + b) % MOD;
  if (a < 0) a += MOD;
}
int n, q;
int a[MAX_N], w[MAX_N];
class FenWickTree {
 public:
  void update(int x, int val) {
    for (int t = x; t < MAX_N; t += t & -t) {
      L[t] += val;
      addmod(sum_weight[t], 1ll * val * a[x]);
      addmod(sum_level[t], 1ll * val * x);
    }
    addmod(::w[x], val);
  }
  long long get(int x) {
    long long res = 0;
    for (int t = x; t > 0; t -= t & -t) {
      res += L[t];
    }
    return res;
  }
  int get_sum_weight(int x) {
    int res = 0;
    for (int t = x; t > 0; t -= t & -t) {
      addmod(res, sum_weight[t]);
    }
    return res;
  }
  int get_weight_range(int u, int v) {
    int res = get_sum_weight(v);
    addmod(res, -get_sum_weight(u - 1));
    return res;
  }
  int get_sum_level(int u, int v) {
    int res = 0;
    for (int t = v; t > 0; t -= t & -t) {
      addmod(res, sum_level[t]);
    }
    for (int t = u - 1; t > 0; t -= t & -t) {
      addmod(res, -sum_level[t]);
    }
    return res;
  }
  long long get_range(int u, int v) { return get(v) - get(u - 1); }

 private:
  static const int MAX_N = ::MAX_N;
  long long L[MAX_N];
  int sum_weight[MAX_N];
  int sum_level[MAX_N];
} fenwickTree;
int get_cost_left(int u, int v) {
  if (u > v) return 0;
  int res = (fenwickTree.get_weight_range(u, v) -
             fenwickTree.get_range(u, v) % INF * a[u]) %
            INF;
  addmod(res, (fenwickTree.get_range(u, v) * 1ll * u -
               fenwickTree.get_sum_level(u, v)));
  return res;
}
int get_cost_right(int u, int v) {
  if (u > v) return 0;
  int res = (fenwickTree.get_range(u, v) % INF * a[v] -
             fenwickTree.get_weight_range(u, v)) %
            INF;
  addmod(res, -((fenwickTree.get_range(u, v) * 1ll * v -
                 fenwickTree.get_sum_level(u, v))));
  return res;
}
int calc(int u, int v) {
  if (u == v) return 0;
  long long total_sum = fenwickTree.get_range(u, v);
  int lo = u, hi = v - 1, res = u;
  while (lo <= hi) {
    int mid = (lo + hi) >> 1;
    long long tmp = fenwickTree.get_range(u, mid);
    if (tmp <= total_sum - tmp) {
      res = mid + 1;
      lo = mid + 1;
    } else
      hi = mid - 1;
  }
  int found = res;
  static const int range = 2;
  for (int j = -range; j <= range; ++j) {
    int cur = res + j;
    if (cur >= u && cur <= v) {
      bool check = true;
      if (cur - 1 >= u)
        check &=
            fenwickTree.get_range(u, cur - 1) <= fenwickTree.get_range(cur, v);
      if (cur + 1 <= v)
        check &=
            fenwickTree.get_range(u, cur) >= fenwickTree.get_range(cur + 1, v);
      if (check) {
        found = cur;
      }
    }
  }
  return found;
}
int calc_cost(int u, int v) {
  if (u == v) return 0;
  int found = calc(u, v);
  int ans = get_cost_right(u, found);
  addmod(ans, get_cost_left(found, v));
  return ans;
}
void solve() {
  cin >> n >> q;
  for (int i = (1), _b = (n + 1); i < _b; ++i) cin >> a[i];
  for (int i = (1), _b = (n + 1); i < _b; ++i) {
    int tmp;
    cin >> tmp;
    fenwickTree.update(i, tmp);
    assert(w[i] == tmp);
  }
  while (q-- > 0) {
    int u, v;
    cin >> u >> v;
    if (u < 0) {
      u = -u;
      fenwickTree.update(u, v - w[u]);
    } else {
      cout << calc_cost(u, v) << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}