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 MOD = (int)1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline unsigned long long icbrt(unsigned long long k) {
  unsigned long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
const int maxn = 2e5 + 5;
int n, q;
int a[maxn];
int w[maxn];
long long fen[maxn];
void upd(int p, int val) {
  p++;
  for (; p < maxn; p += p & -p) {
    fen[p] += val;
  }
}
long long query(int p) {
  p++;
  long long res = 0;
  for (; p > 0; p -= p & -p) {
    res += fen[p];
  }
  return res;
}
int fen2[2][maxn];
void upd(int p, int fen[], int val) {
  p++;
  for (; p < maxn; p += p & -p) {
    addmod(fen[p], val);
  }
}
int query(int fen[], int p) {
  p++;
  int res = 0;
  for (; p > 0; p -= p & -p) {
    addmod(res, fen[p]);
  }
  return res;
}
int query(int fen[], int l, int r) {
  int res = query(fen, r);
  submod(res, query(fen, l - 1));
  return res;
}
void volamtruyenkyii() {
  cin >> n >> q;
  for (int i = (0); i < (n); ++i) cin >> a[i], submod(a[i], i);
  for (int i = (0); i < (n); ++i) {
    cin >> w[i];
    upd(i, w[i]);
    upd(i, fen2[0], w[i]);
    upd(i, fen2[1], mult(a[i], w[i]));
  }
  while (q--) {
    int u, v;
    cin >> u >> v;
    if (u < 0) {
      u = -u - 1;
      upd(u, -w[u]);
      upd(u, fen2[0], (MOD - w[u]) % MOD);
      upd(u, fen2[1], (MOD - mult(a[u], w[u])) % MOD);
      w[u] = v;
      upd(u, w[u]);
      upd(u, fen2[0], w[u]);
      upd(u, fen2[1], mult(a[u], w[u]));
    } else {
      u--, v--;
      long long sum = query(v) - query(u - 1);
      int l = u, r = v;
      while (l < r) {
        int m = l + r >> 1;
        if (2 * (query(m) - query(u - 1)) < sum) {
          l = m + 1;
        } else {
          r = m;
        }
      }
      int m = l + r >> 1;
      int x = a[m];
      int res = 0;
      addmod(res, mult(x, query(fen2[0], u, m)));
      submod(res, query(fen2[1], u, m));
      addmod(res, query(fen2[1], m + 1, v));
      submod(res, mult(x, query(fen2[0], m + 1, v)));
      cout << res << "\n";
    }
  }
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  volamtruyenkyii();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}