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;
int n, m;
long long seg[400005], lazy[400005];
set<pair<pair<int, int>, long long> > s;
void down(int node, int a, int b) {
  if (lazy[node] != 0) {
    seg[node] += lazy[node] * (b - a + 1);
    if (a != b) {
      lazy[2 * node] += lazy[node];
      lazy[2 * node + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
}
void update(int node, int a, int b, int x, int y, long long val) {
  down(node, a, b);
  if (b < x || a > y) {
    return;
  }
  if (a >= x && b <= y) {
    seg[node] += val * (b - a + 1);
    if (a != b) {
      lazy[2 * node] += val;
      lazy[2 * node + 1] += val;
    }
    return;
  }
  int mid = (a + b) / 2;
  update(2 * node, a, mid, x, y, val);
  update(2 * node + 1, mid + 1, b, x, y, val);
  seg[node] = seg[2 * node] + seg[2 * node + 1];
}
long long query(int node, int a, int b, int x, int y) {
  down(node, a, b);
  if (b < x || a > y) {
    return 0;
  }
  if (a >= x && b <= y) {
    return seg[node];
  }
  int mid = (a + b) / 2;
  return query(2 * node, a, mid, x, y) + query(2 * node + 1, mid + 1, b, x, y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    s.insert(make_pair(make_pair(i, i), i));
  }
  for (int i = 1; i <= m; i++) {
    long long type, l, r, x;
    cin >> type >> l >> r;
    if (type == 1) {
      cin >> x;
      auto f = s.lower_bound(make_pair(make_pair(l, 0), -1));
      if (f != s.begin()) {
        f--;
        if ((*f).first.second != l - 1) {
          s.insert(make_pair(make_pair((*f).first.first, l - 1), (*f).second));
          s.insert(make_pair(make_pair(l, (*f).first.second), (*f).second));
          s.erase(f);
        }
      }
      auto b = s.upper_bound(make_pair(make_pair(r, n + 1), 1 << 30));
      b--;
      if ((*b).first.second != r) {
        s.insert(make_pair(make_pair((*b).first.first, r), (*b).second));
        s.insert(make_pair(make_pair(r + 1, (*b).first.second), (*b).second));
        s.erase(b);
      }
      auto it = s.lower_bound(make_pair(make_pair(l, 0), -1));
      while (it != s.end() && (*it).first.second <= r) {
        auto curr = it;
        it++;
        update(1, 1, n, (*curr).first.first, (*curr).first.second,
               abs((*curr).second - x));
        s.erase(curr);
      }
      s.insert(make_pair(make_pair(l, r), x));
    } else {
      cout << query(1, 1, n, l, r) << "\n";
    }
  }
}
