DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, neg = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * neg;
}
inline void print(long long x) {
  if (x < 0) {
    putchar('-');
    print(abs(x));
    return;
  }
  if (x <= 9)
    putchar(x + '0');
  else {
    print(x / 10);
    putchar(x % 10 + '0');
  }
}
inline long long qpow(long long x, long long e, long long _MOD) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * x % _MOD;
    x = x * x % _MOD;
    e >>= 1;
  }
  return ans;
}
struct SegTree {
  struct node {
    long long l, r, val, sum, lz;
    bool is_same;
  } s[100005 << 2];
  inline void pushup(long long k) {
    if (s[k << 1].is_same && s[k << 1 | 1].is_same &&
        s[k << 1].val == s[k << 1 | 1].val) {
      s[k].is_same = 1;
      s[k].val = s[k << 1].val;
    } else {
      s[k].is_same = 0;
      s[k].val = 0;
    }
    s[k].sum = s[k << 1].sum + s[k << 1 | 1].sum;
  }
  inline void pushdown(long long k) {
    if (s[k].lz) {
      s[k << 1].is_same = 1;
      s[k << 1].lz += s[k].lz;
      s[k << 1].sum += (s[k << 1].r - s[k << 1].l + 1) * s[k].lz;
      s[k << 1].val = s[k].val;
      s[k << 1 | 1].is_same = 1;
      s[k << 1 | 1].lz += s[k].lz;
      s[k << 1 | 1].sum += (s[k << 1 | 1].r - s[k << 1 | 1].l + 1) * s[k].lz;
      s[k << 1 | 1].val = s[k].val;
      s[k].lz = 0;
    }
  }
  inline void build(long long k, long long l, long long r) {
    s[k].l = l;
    s[k].r = r;
    if (l == r) {
      s[k].is_same = 1;
      s[k].val = l;
      return;
    }
    long long mid = (l + r) >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
    pushup(k);
  }
  inline void modify(long long k, long long l, long long r, long long x) {
    if (l <= s[k].l && s[k].r <= r && s[k].is_same) {
      s[k].lz += abs(s[k].val - x);
      s[k].sum += (s[k].r - s[k].l + 1) * abs(s[k].val - x);
      s[k].val = x;
      return;
    }
    pushdown(k);
    long long mid = (s[k].l + s[k].r) >> 1;
    if (r <= mid)
      modify(k << 1, l, r, x);
    else if (l > mid)
      modify(k << 1 | 1, l, r, x);
    else
      modify(k << 1, l, mid, x), modify(k << 1 | 1, mid + 1, r, x);
    pushup(k);
  }
  inline long long query(long long k, long long l, long long r) {
    if (l <= s[k].l && s[k].r <= r) {
      return s[k].sum;
    }
    pushdown(k);
    long long mid = (s[k].l + s[k].r) >> 1;
    if (r <= mid)
      return query(k << 1, l, r);
    else if (l > mid)
      return query(k << 1 | 1, l, r);
    else
      return query(k << 1, l, mid) + query(k << 1 | 1, mid + 1, r);
  }
} st;
long long n = read(), m = read();
signed main() {
  st.build(1, 1, n);
  while (m--) {
    long long opt = read();
    if (opt == 1) {
      long long x = read(), y = read(), c = read();
      st.modify(1, x, y, c);
    } else {
      long long x = read(), y = read();
      cout << st.query(1, x, y) << endl;
    }
  }
  return 0;
}
