diff --git a/SM2-WWMM-(2).md b/SM2-WWMM-(2).md
index 863f5ee..0317445 100644
--- a/SM2-WWMM-(2).md
+++ b/SM2-WWMM-(2).md
@@ -399,7 +399,6 @@ $t_0=0+0$
 ### 方案二：(移位、加法、减法)
 因为Order素数不是MFMM，所以这个方案其实优势不大、甚至没有优势，尤其是在使用**使用MULXQ/ADCXQ/ADOXQ**的情况下。  
 移位针对 $O_3$ $O_2$ 乘法  
-这里加法没有溢出风险：Y不可能是0xFFFFFFFFFFFFFFFF。  
 
 $T_2=T_1 \ast O=Y \ast O= Y \ast 2^{256}-(Y \ast 2^{32}) \ast 2^{192} - Y \ast 2^{128} + (Y \ast O_1) \ast 2^{64} + (Y \ast O_0)$  
 
@@ -407,12 +406,11 @@ $T_3=T + T_2=t_7 \ast 2^{448} + t_6 \ast 2^{384} + t_5 \ast 2^{320} + t_4 \ast 2
 $T_3=t_7 \ast 2^{448} + t_6 \ast 2^{384} + t_5 \ast 2^{320} + (t_4+Y) \ast 2^{256}+(t_3 - Y \ast 2^{32}) \ast 2^{192} + (t_2 - Y) \ast 2^{128} + (t_1 + Y \ast O_1) \ast 2^{64} + (t_0 + Y \ast O_0)  $  
 
 ```asm
-	// First reduction step, [ord3, ord2, ord1, ord0] = [1, -0x100000000, -1, ord1, ord0]
+	// First reduction step
 	MOVQ acc0, AX
 	MULQ p256ordK0<>(SB)
 	MOVQ AX, t0                 // Y = t0 = (k0 * acc0) mod 2^64
-	// calculate the positive part first: [1, 0, 0, ord1, ord0] * t0 + [0, acc3, acc2, acc1, acc0]
-	// the result is [acc0, acc3, acc2, acc1], last lowest limb is dropped.
+
 	MOVQ p256ord<>+0x00(SB), AX
 	MULQ t0
 	ADDQ AX, acc0               // (carry1, acc0) = acc0 + L(t0 * ord0)
@@ -420,6 +418,16 @@ $T_3=t_7 \ast 2^{448} + t_6 \ast 2^{384} + t_5 \ast 2^{320} + (t_4+Y) \ast 2^{25
 	MOVQ DX, t1                 // t1 = carry1 + H(t0 * ord0)
 	MOVQ t0, acc0               // acc0 =  t0
 
+	// calculate the negative part: [acc0, acc3, acc2, acc1] - [0, 0x100000000, 1, 0] * t0
+	MOVQ t0, AX
+	MOVQ t0, DX
+	SHLQ $32, AX
+	SHRQ $32, DX
+
+	SUBQ t0, acc2
+	SBBQ AX, acc3
+	SBBQ DX, acc0
+
 	MOVQ p256ord<>+0x08(SB), AX
 	MULQ t0
 	ADDQ t1, acc1               // (carry2, acc1) = acc1 + t1
@@ -429,15 +437,6 @@ $T_3=t_7 \ast 2^{448} + t_6 \ast 2^{384} + t_5 \ast 2^{320} + (t_4+Y) \ast 2^{25
 	ADCQ DX, acc2
 	ADCQ $0, acc3
 	ADCQ $0, acc0
-	// calculate the negative part: [acc0, acc3, acc2, acc1] - [0, 0x100000000, 1, 0] * t0
-	MOVQ t0, AX
-	MOVQ t0, DX
-	SHLQ $32, AX
-	SHRQ $32, DX
-
-	SUBQ t0, acc2
-	SBBQ AX, acc3
-	SBBQ DX, acc0
 
 ```
 乘法: 3  
@@ -611,16 +610,7 @@ $t_5=t_5 - 0$
 	ADCQ $0, DX
 	MOVQ DX, BX
 
-	MOVQ p256ord<>+0x08(SB), AX
-	MULQ t0
-	ADDQ BX, acc1
-	ADCQ $0, DX
-	ADDQ AX, acc1
-	ADCQ DX, acc2
-	ADCQ $0, acc3
-	ADCQ t0, acc4
-	ADCQ $0, acc5
-
+	MOVQ t0, acc0
 	MOVQ t0, AX
 	MOVQ t0, DX
 	SHLQ $32, AX
@@ -628,8 +618,17 @@ $t_5=t_5 - 0$
 		
 	SUBQ t0, acc2
 	SBBQ AX, acc3
-	SBBQ DX, acc4
-	SBBQ $0, acc5
+	SBBQ DX, acc0
+
+	MOVQ p256ord<>+0x08(SB), AX
+	MULQ t0
+	ADDQ BX, acc1
+	ADCQ $0, DX
+	ADDQ AX, acc1
+	ADCQ DX, acc2
+	ADCQ $0, acc3
+	ADCQ acc0, acc4
+	ADCQ $0, acc5
 ```
 乘法: 3  
 移位：2    
@@ -674,7 +673,7 @@ $t_5=t_5 - 0$
 | ----------- | ----------- | ----------- | ----------- | ----------- |
 | 方案一      | 5 | 0 | 15 | 0 |
 | 方案一（MULX/ADCX/ADOX）   | 5 | 0 | 10 | 0 |
-| 方案二   | 3 | 2 | 9 | 4 |  
+| 方案二   | 3 | 2 | 9 | 3 |  
 | 方案二（MULX）   | 3 | 2 | 8 | 4 |  
 
 看来在支持**MULXQ/ADCXQ/ADOXQ**的情况下，使用方案一（MULX/ADCX/ADOX）更好！