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 int N = 400005;
long long tree[N], lazy[N];
void updateRange(int node, int start, int end, int l, int r, long long val) {
  if (lazy[node] != 0) {
    tree[node] += (end - start + 1) * lazy[node];
    if (start != end) {
      lazy[node * 2] += lazy[node];
      lazy[node * 2 + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (start > end or start > r or end < l) return;
  if (start >= l and end <= r) {
    tree[node] += (end - start + 1) * val;
    if (start != end) {
      lazy[node * 2] += val;
      lazy[node * 2 + 1] += val;
    }
    return;
  }
  int mid = (start + end) / 2;
  updateRange(node * 2, start, mid, l, r, val);
  updateRange(node * 2 + 1, mid + 1, end, l, r, val);
  tree[node] = tree[node * 2] + tree[node * 2 + 1];
}
long long queryRange(int node, int start, int end, int l, int r) {
  if (start > end or start > r or end < l) return 0;
  if (lazy[node] != 0) {
    tree[node] += (end - start + 1) * lazy[node];
    if (start != end) {
      lazy[node * 2] += lazy[node];
      lazy[node * 2 + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (start >= l and end <= r) return tree[node];
  int mid = (start + end) / 2;
  long long p1 = queryRange(node * 2, start, mid, l, r);
  long long p2 = queryRange(node * 2 + 1, mid + 1, end, l, r);
  return (p1 + p2);
}
int n, m;
set<pair<int, pair<int, int> > > st;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) st.insert(make_pair(i, make_pair(i, i)));
  while (m--) {
    int tp, l, r, x;
    scanf("%d", &tp);
    if (tp == 1) {
      scanf("%d %d %d", &l, &r, &x);
      auto it = st.lower_bound(make_pair(l, make_pair(0, 0)));
      pair<int, pair<int, int> > A, B;
      vector<pair<int, pair<int, int> > > vec;
      A = *it;
      A.first = l - 1;
      while (it != st.end() and it->second.first <= r) {
        int L = max(l, it->second.first);
        int R = min(r, it->first);
        int V = (((it->second.second) - x) < 0 ? -((it->second.second) - x)
                                               : ((it->second.second) - x));
        updateRange(1, 1, n, L, R, V);
        B = *it;
        B.second.first = r + 1;
        vec.push_back(*it);
        it++;
      }
      for (int i = 0; i < vec.size(); i++) st.erase(vec[i]);
      if (A.second.first <= A.first) st.insert(A);
      if (B.second.first <= B.first) st.insert(B);
      st.insert(make_pair(r, make_pair(l, x)));
    } else {
      scanf("%d %d", &l, &r);
      long long ans = queryRange(1, 1, n, l, r);
      printf("%I64d\n", ans);
    }
  }
}
