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;
template <typename T>
inline void ckmax(T& x, T y) {
  x = (y > x ? y : x);
}
template <typename T>
inline void ckmin(T& x, T y) {
  x = (y < x ? y : x);
}
namespace Fread {
const int SIZE = 1 << 21;
char buf[SIZE], *S, *T;
inline char getchar() {
  if (S == T) {
    T = (S = buf) + fread(buf, 1, SIZE, stdin);
    if (S == T) return '\n';
  }
  return *S++;
}
}  // namespace Fread
namespace Fwrite {
const int SIZE = 1 << 21;
char buf[SIZE], *S = buf, *T = buf + SIZE;
inline void flush() {
  fwrite(buf, 1, S - buf, stdout);
  S = buf;
}
inline void putchar(char c) {
  *S++ = c;
  if (S == T) flush();
}
struct NTR {
  ~NTR() { flush(); }
} ztr;
}  // namespace Fwrite
namespace Fastio {
struct Reader {
  template <typename T>
  Reader& operator>>(T& x) {
    char c = Fread ::getchar();
    T f = 1;
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = Fread ::getchar();
    }
    x = 0;
    while (c >= '0' && c <= '9') {
      x = x * 10 + (c - '0');
      c = Fread ::getchar();
    }
    x *= f;
    return *this;
  }
  Reader& operator>>(char& c) {
    c = Fread ::getchar();
    while (c == '\n' || c == ' ') c = Fread ::getchar();
    return *this;
  }
  Reader& operator>>(char* str) {
    int len = 0;
    char c = Fread ::getchar();
    while (c == '\n' || c == ' ') c = Fread ::getchar();
    while (c != '\n' && c != ' ') {
      str[len++] = c;
      c = Fread ::getchar();
    }
    str[len] = '\0';
    return *this;
  }
  Reader() {}
} cin;
const char endl = '\n';
struct Writer {
  template <typename T>
  Writer& operator<<(T x) {
    if (x == 0) {
      Fwrite ::putchar('0');
      return *this;
    }
    if (x < 0) {
      Fwrite ::putchar('-');
      x = -x;
    }
    static int sta[45];
    int top = 0;
    while (x) {
      sta[++top] = x % 10;
      x /= 10;
    }
    while (top) {
      Fwrite ::putchar(sta[top] + '0');
      --top;
    }
    return *this;
  }
  Writer& operator<<(char c) {
    Fwrite ::putchar(c);
    return *this;
  }
  Writer& operator<<(char* str) {
    int cur = 0;
    while (str[cur]) Fwrite ::putchar(str[cur++]);
    return *this;
  }
  Writer& operator<<(const char* str) {
    int cur = 0;
    while (str[cur]) Fwrite ::putchar(str[cur++]);
    return *this;
  }
  Writer() {}
} cout;
}  // namespace Fastio
const int MAXN = 2e5;
const int MOD = 1e9 + 7;
int n, q, a[MAXN + 5], w[MAXN + 5];
struct FenwickTree {
  __int128 sum[MAXN + 5];
  void point_add(int p, __int128 v) {
    for (int i = p; i <= n; i += (i & (-i))) {
      sum[i] += v;
    }
  }
  __int128 presum_query(int p) {
    __int128 res = 0;
    for (int i = p; i; i -= (i & (-i))) {
      res += sum[i];
    }
    return res;
  }
  __int128 range_sum_query(int l, int r) {
    if (l > r) {
      return 0;
    }
    return presum_query(r) - presum_query(l - 1);
  }
  FenwickTree() {}
};
FenwickTree T, T1, T2;
__int128 calc(int l, int r, int i) {
  return T1.range_sum_query(l, i - 1) * (a[i] - i) -
         T2.range_sum_query(l, i - 1) +
         T1.range_sum_query(i + 1, r) * (i - a[i]) +
         T2.range_sum_query(i + 1, r);
}
int main() {
  Fastio ::cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    Fastio ::cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    Fastio ::cin >> w[i];
    T1.point_add(i, w[i]);
    T2.point_add(i, (__int128)w[i] * (a[i] - i));
  }
  while (q--) {
    int x, y;
    Fastio ::cin >> x >> y;
    if (x < 0) {
      x = -x;
      __int128 d = (y - w[x]);
      T1.point_add(x, d);
      T2.point_add(x, d * (a[x] - x));
      w[x] = y;
    } else {
      int l = x, r = y;
      while (l < r) {
        int mid = (l + r) >> 1;
        if (T1.range_sum_query(x, mid) < T1.range_sum_query(mid + 1, y)) {
          l = mid + 1;
        } else {
          r = mid;
        }
      }
      __int128 ans = calc(x, y, l);
      if (l > x) {
        ckmin(ans, calc(x, y, l - 1));
      }
      Fastio ::cout << ans % MOD << Fastio ::endl;
    }
  }
  return 0;
}