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 double pi = acos(-1.0);
const double eps = 1e-11;
template <class T>
inline void ckmin(T& a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void ckmax(T& a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const int maxn = (1 << 20);
const int MOD = 1000000007;
template <typename base_type, base_type _MOD>
class IntMod {
 public:
  static const int INVERSE_CACHE_SIZE = (1 << 20);
  static base_type MOD;
  static void set_mod(base_type new_mod) { MOD = new_mod; }
  base_type n;
  IntMod(long long d = 0) { n = (d >= 0 ? d % MOD : (d % MOD + MOD) % MOD); }
  virtual ~IntMod() = default;
  IntMod operator-() const { return Build(n == 0 ? 0 : MOD - n); }
  IntMod& operator+=(IntMod a) {
    n = (n >= MOD - a.n ? n - MOD + a.n : n + a.n);
    return *this;
  }
  IntMod& operator-=(IntMod a) {
    n = (n >= a.n) ? n - a.n : n - a.n + MOD;
    return *this;
  }
  IntMod& operator*=(IntMod a) {
    *this = *this * a;
    return *this;
  }
  IntMod& operator/=(IntMod a) {
    *this = *this / a;
    return *this;
  }
  static IntMod build(base_type n) {
    IntMod r;
    r.n = n;
    return r;
  }
  static base_type inverse_cache[INVERSE_CACHE_SIZE];
  static bool inverse_cache_ready;
  friend IntMod inverse(IntMod n) { return build(inverse_internal(n.n)); }
  static base_type inverse_internal(base_type n) {
    if (!inverse_cache_ready) {
      inverse_cache_ready = true;
      inverse_cache[0] = 0;
      inverse_cache[1] = 1;
      for (int n = 2; n < INVERSE_CACHE_SIZE; ++n)
        inverse_cache[n] =
            (MOD -
             (base_type)((long long)inverse_cache[MOD % n] * (MOD / n) % MOD));
    }
    return n < INVERSE_CACHE_SIZE
               ? inverse_cache[n]
               : MOD - (base_type)((long long)inverse_internal(MOD % n) *
                                   (MOD / n) % MOD);
  }
  friend bool operator==(IntMod a, IntMod b) { return a.n == b.n; }
  friend bool operator!=(IntMod a, IntMod b) { return a.n != b.n; }
  friend IntMod operator+(IntMod a, IntMod b) {
    return build(a.n >= MOD - b.n ? a.n - MOD + b.n : a.n + b.n);
  }
  friend IntMod operator-(IntMod a, IntMod b) {
    return build(a.n >= b.n ? a.n - b.n : a.n - b.n + MOD);
  }
  friend IntMod operator*(IntMod a, IntMod b) {
    return build(
        static_cast<base_type>(static_cast<long long>(a.n) * b.n % MOD));
  }
  friend IntMod operator/(IntMod a, IntMod b) { return a * inverse(b); }
  friend IntMod pow(IntMod p, long long e) {
    if (e <= 0) return IntMod(1);
    IntMod r = IntMod(1);
    while (1) {
      if (e & 1) r *= p;
      e /= 2;
      if (e)
        p = p * p;
      else
        break;
    }
    return r;
  }
  friend istream& operator>>(istream& stream, IntMod& a) {
    stream >> a.n;
    return stream;
  }
  friend ostream& operator<<(ostream& stream, const IntMod& a) {
    stream << a.n;
    return stream;
  }
};
template <typename base_type, base_type _MOD>
base_type IntMod<base_type, _MOD>::inverse_cache[INVERSE_CACHE_SIZE];
template <typename base_type, base_type _MOD>
bool IntMod<base_type, _MOD>::inverse_cache_ready;
template <typename base_type, base_type _MOD>
base_type IntMod<base_type, _MOD>::MOD = _MOD;
using Int = IntMod<int, MOD>;
int n;
int a[maxn], w[maxn];
Int sw[maxn], swa[maxn];
long long sw2[maxn];
void update_sa(Int s[], int p, Int delta) {
  for (++p; p <= n; p = (p | (p - 1)) + 1) s[p] += delta;
}
Int get_sa(Int s[], int p) {
  Int r = 0;
  for (; p > 0; p &= (p - 1)) r += s[p];
  return r;
}
Int calc(Int s[], int a, int b) {
  if (a > b) return 0;
  return get_sa(s, b + 1) - get_sa(s, a);
}
void update_sa2(long long s[], int p, long long delta) {
  for (++p; p <= n; p = (p | (p - 1)) + 1) s[p] += delta;
}
long long get_sa2(long long s[], int p) {
  long long r = 0;
  for (; p > 0; p &= (p - 1)) r += s[p];
  return r;
}
long long calc2(long long s[], int a, int b) {
  if (a > b) return 0;
  return get_sa2(s, b + 1) - get_sa2(s, a);
}
void update_weight(int id, int new_w) {
  Int delta = new_w - w[id];
  update_sa(sw, id, delta);
  update_sa(swa, id, delta * a[id]);
  update_sa2(sw2, id, new_w - w[id]);
  w[id] = new_w;
}
int query(int s, int t) {
  int low = s, high = t, m;
  while (1) {
    m = (low + high) / 2;
    long long s1 = calc2(sw2, s, m - 1);
    long long s2 = calc2(sw2, m + 1, t);
    if (s1 + w[m] < s2) {
      low = m + 1;
      continue;
    }
    if (s2 + w[m] < s1) {
      high = m - 1;
      continue;
    }
    break;
  }
  Int ret = 0;
  ret += calc(sw, s, m) * a[m];
  ret -= calc(swa, s, m);
  ret += calc(swa, m + 1, t);
  ret -= calc(sw, m + 1, t) * a[m];
  return ret.n;
}
int main() {
  std::ios::sync_with_stdio(false);
  int nq;
  while (cin >> n >> nq) {
    for (int i = 0; i < (n); ++i) cin >> a[i];
    for (int i = 0; i < (n); ++i) cin >> w[i];
    for (int i = 0; i < (n); ++i) a[i] -= i;
    for (int i = 0; i < (n + 1); ++i) sw[i] = swa[i] = 0;
    for (int i = 0; i < (n + 1); ++i) sw2[i] = 0;
    for (int i = 0; i < (n); ++i) update_sa(sw, i, w[i]);
    for (int i = 0; i < (n); ++i)
      update_sa(swa, i, (long long)w[i] * a[i] % MOD);
    for (int i = 0; i < (n); ++i) update_sa2(sw2, i, w[i]);
    for (int cl = 0; cl < (nq); ++cl) {
      int x, y;
      cin >> x >> y;
      if (x < 0)
        update_weight(-x - 1, y);
      else {
        int r = query(x - 1, y - 1);
        printf("%d\n", r);
      }
    }
    break;
  }
  return 0;
}