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;
const long long maxn = 4e6 + 10, mOd = 1e9 + 7, inf = 2e9;
long long seg[maxn], same[maxn], lzy[maxn], n;
inline void build(long long id, long long l, long long r) {
  if (r - l == 1) {
    same[id] = l + 1;
    return;
  }
  long long left = (id << 1), right = left | 1, mid = (l + r) >> 1;
  build(left, l, mid);
  build(right, mid, r);
  return;
}
inline void shift(long long id, long long l, long long r) {
  if (!lzy[id]) return;
  if (r - l <= 1) {
    return;
  }
  long long left = (id << 1), right = left | 1, mid = (l + r) >> 1;
  if (same[id]) same[left] = same[right] = same[id];
  lzy[left] += lzy[id];
  lzy[right] += lzy[id];
  seg[left] += (mid - l) * lzy[id];
  seg[right] += (r - mid) * lzy[id];
  lzy[id] = 0;
  return;
}
inline void update(long long id, long long low, long long high, long long l,
                   long long r, long long val) {
  if (low >= r || high <= l) {
    return;
  }
  if (low <= l && high >= r && same[id]) {
    shift(id, l, r);
    lzy[id] += abs(val - same[id]);
    seg[id] += abs(val - same[id]) * (r - l);
    same[id] = val;
    return;
  }
  long long left = (id << 1), right = left | 1, mid = (l + r) >> 1;
  shift(id, l, r);
  update(left, low, high, l, mid, val);
  update(right, low, high, mid, r, val);
  if (same[left] == same[right]) {
    same[id] = same[left];
  } else
    same[id] = 0;
  seg[id] = seg[left] + seg[right];
  return;
}
inline long long query(long long id, long long low, long long high, long long l,
                       long long r) {
  if (low >= r || high <= l) {
    return 0;
  }
  if (low <= l && high >= r) {
    return seg[id];
  }
  long long left = (id << 1), right = left | 1, mid = (l + r) >> 1;
  shift(id, l, r);
  return query(left, low, high, l, mid) + query(right, low, high, mid, r);
}
int main() {
  long long m;
  cin >> n >> m;
  build(1, 0, n);
  while (m--) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long x, y, val;
      cin >> x >> y >> val;
      update(1, x - 1, y, 0, n, val);
    } else {
      long long x, y;
      cin >> x >> y;
      cout << query(1, x - 1, y, 0, n) << endl;
    }
  }
}
